Dont copié formé thé book oh ya dont copié formé thé book cause you Oiil inde up being triste soi remembré not toi copié frome thé book oh ya!translation in english please!
Helle, i need help for simplify the expression: [( cos x) ( sin x) - ( sin x) ( - sin x)] / ( sin x) ²
Solve the following equation for 0 less than and/or equal to "x" less than and/or equal to 360--cos^2x - 1 = sin^2x--Attempt:cos^2x - 1 - sin^2x = 0cos^2x - 1 - (1 - cos^2x) = 0cos^2x - 1 - 1 + cos^2x = 02cos^2x - 2 = 0(2cos^2x/2)= (-2/2)cos^2x = -1cosx = square root of -1And I c...
cos(tan + cot) = csconly simplify one side to equal cscso far I got this far: [((cos)(sin))/(cos)] + [((cos)(cos))/(sin)] = cscI don't know what to do next
the original problem was: (sin x + cos x)^2 + (sin x - cos x)^2 = 2steps too pleaseI got 1 for (sin x + cos x)^2 but then what does (sin x - cos x)^2 become since it's minus?
The identity of: sin^4x+cos^4x= (sin^2x+cos^2x)(sin^2x+cos^2x) sin^2x+cos^2=1 This is the answer I come up with, it is not one of the options available as an answer. The answers given are 1. -2sin^2xcos^2x 2. 1+2sin^2x-2sin^4x 3. 1+3sin^3x-2sin^2x 4. 1-2sin^2x+2sin^4x 5. 0
find all solutions in the interval [0,2 pi) sin(x+(3.14/3) + sin(x- 3.14/3) =1 sin^4 x cos^2 xSince sin (a+b) = sina cosb + cosb sina and sin (a-b) = sina cosb - cosb sina, the first problem can be written 2 sin x cos (pi/3)= sin x The solution to sin x = 1 is x = pi/2 For your o...
2) Use the sum and difference identites sin[x + pi/4] + sin[x-pi/4] = -1 sinx cospi/4 + cosxsin pi/4 + sinx cos pi/4 - cosx sin pi/4 = -1 2 sin x cos pi/4 =-1 cos pi/4 = sqr2/2 2sin^x(sqrt2/2) = -1 sin x = -sqrt2 x = 7pi/4 and 5pi/4 Am I correct?
sin^2 theta + 2 cos theta -1/ (sin^2 theta +3 cos theta-3= cos^2 theta+cos theta/ (-sin^2 theta)Prove the identity.
Let g(x) = sin (cos x^3) Find g ' (x):The choices are a) -3x^2sinx^3cos(cos x^3)b) -3x^2sinx^3sin(cos x^3)c) -3x^2cosx^3sin(cos x^3)d) 3x^2sin^2(cos x^3)I'm not exactly sure where I should start.Should I begin with d/dx of sin? Or do the inside derivative first...and do I have t...
the original problem was:Solve: sin(3x)-sin(x)=cos(2x)so far i've gooten to:sin(x)(2sin(x)cos(x)-1)=cos^2(x)-sin^2(x)Where would I go from here?
the original problem was: Solve: sin(3x)-sin(x)=cos(2x) so far i've gotten to: sin(x)(2sin(x)cos(x)-1)=cos^2(x)-sin^2(x) Where would I go from here?